Determine the change in enthalpy and internal energy when 1 mole of water completely vaporises at `100^(@)C` and 1 atm pressure. [Latent heat of vaporisation of water=536`cal*g^(-1)`]

Calculate the work done when 1 mol of water vaporises at 100^(@)C and 1 atm pressure. Assume water vapour behaves like an ideal gas.

Calculate the values of Delta H and DeltaU in the vaporisation of 90g of water at 100^(@)C and 1 atm pressure. The latent heat of vaporisation of water at the same temperature and pressure =540cal*g^(-1)

Calculate the K_b for water given latent heat of vaporisation of water 540 Cal//gm .

Calculate the free energy change and entropy change per mole when liquid water boils at 1 atmosphere latent heat of vaporization of water= 2.0723KJ mol^(-1)

Assuming that water vapour is an ideal gas, the internal energy change (DeltaU) when 1 mol of water is vaporised at 1 bar pressure and 100^(@)C , will be (Given: at 1 bar and 373 K, molar enthalpy of vaporisation of water is 41kJ*mol^(-1),R=8.3J*mol^(-1)*K^(-1) )-

The change in enthalpy of vaporisation of 1 mol of water is 40850 " J mol"^(-1) " at " 100^(@)C and 1 atm pressure. What is the difference between molar entropy of water and water vapour at that temperature and pressure ?

Calculate (1) the enthalpy change in the fusion of 100 g ice at 0^(@)C temperature and 1 atm pressure pressure (2) the enthalpy change in the vaporisation of 10g water at 100^(@)C temperature and 1 atm pressure. Givenn : Latent heat of ice at 0^(@)C temperature and 1 atm pressure =6.02kJ*mol^(-1) and latent heat of vaporisation of water at 100^(@)C temperature= 40.4kJ*mol^(-1) .

Heat required to vaporise 1 g of water at 100^(@)C and 1 atm is 2.26 kJ. What is the enthalpy of vaporisation of water at this temperature and pressure?

169 J energy is required to transform 1g (1cm^(3)) of water into steam at 1 atm pressure. If the latent heat of vaporisation of water is 540 cal cdot g^(-1) , the volume of that steam will be